Find all $p,q$ such that $(5p+q) \mid 168$ and $(5p-q) \mid 168$ 
Let $p,q$ with $p < q$ be coprime positive integers. Find all $p,q$ such that $(5p+q) \mid 168$ and $(5p-q) \mid 168$.

I first factorized $168 = 2^3 \cdot 3 \cdot 7$ but didn't see how to use the system of divisibility to continue. How do we use the system of divisibility here in order to solve the question?
 A: Hint: There are $16$ positive divisors of 168.  There are $32$ positive or negative divisors of $168$.  
You can construct the system
\begin{align}
5p+q&=a\\
5p-q&=b
\end{align}
where $a$ is a positive divisor of $168$ and $b$ is a positive or negative divisor.  Solve the simultaneous system for $p$ and $q$, using the fractions which appear to figure out what $a$'s and $b$'s are possible.
A: The (positive) divisors of $168$ are
$$1,3,7,21,\\2,6,14,42,\\4,12,28,84,\\8,24,56,168$$
Let $a$ and $b$ be any of these or their negatives.  If $a=5p+q$ and $b=5p-q$, then $a+b=10p$ and $a-b=2q$, so $a$ and $b$ must have the same parity and sum to a multiple of $10$.  If $p$ and $q$ are coprime, then $\gcd(a+b,a-b)=\gcd(10p,2q)=2\gcd(5p,q)=2$, since $\gcd(p,q)=1$ and $5\not\mid q$ (because $5\not\mid168$).
This leaves a relatively short list of possibilies for $(a,b)$.  Listing them with $a\gt b$, the possibilities are
$$(21,-1),\\(7,3),\\
(12,-2),(28,2),(8,2),(168,2)\\
(6,4),(56,-6),\\
(14,-4),(24,-14)\\
(42,8)$$
(where each row takes a number from the list at top of divisors and picks out all later numbers from that list that satisfy the conditions).  These lead to $(p,q)$ pairs
$$(2,11),\\
(1,2),\\
(1,7),(3,13),(1,3),(17,83),\\
(1,1),(5,31),\\
(1,9),(1,19),\\
(5,17)
$$
The OP's stipulation that $p\lt q$ eliminates $(1,1)$.
